Answer:
Step-by-step explanation:
Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.
The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.
Substituting 114 - 2s for h in the volume formula, we obtain:
V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)
This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:
dV
----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2
ds
Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.
Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in
and the volume is V = s^2(h) = 46,298.25 in^3
Answer:
950
Step-by-step explanation:
2500(0.076)(5) = 950
Let's write an inequality, such as follows: x < sqrt(50) < y. Square both sides of the equation. We get x^2 < 50 < y^2. Obviously, x is between 7 and 8. Also notice, that for integers a,b, (ab)^2/b^2, equals a^2. So let's try values, like 7.1. Using the previous fact, (7.1)^2, equals (71)^2/100. So, (7.1)^2, equals 50.41. Thus, our number is between 7 and 7.1. We find, with a bit of experimentation, that the square root of 50, is 7.07.
Answer:
- (2x + 5)/(x - 9)
- 3. quantity 2 x plus 5 over x minus 9
Step-by-step explanation:
Quantity 2 x squared plus 13 x plus 20 all over x squared minus 5 x minus 36
- (2x² + 13x + 20)/(x² -5x - 36) =
- (2x²+ 5x + 8x + 20)/(x² - 9x + 4x -36)=
- (2x + 5)(x + 4)/(x - 9)(x + 4) =
- (2x + 5)/(x - 9)
Correct option is option 3
Answer:
w = 
Step-by-step explanation:
Using Pythagoras' identity in the right triangle
w² + 7² = 14²
w² + 49 = 196 ( subtract 49 from both sides )
w² = 147 ( take the square root of both sides )
w =
